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Q.

The value of n1C0.1C0+n1C1.nC2+n1C2.nC3++n1Cn1.nCn is equal to 

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a

2nCn

b

2n1Cn1

c

2nCn+1

d

2nCn1

answer is B.

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Detailed Solution

n1C0.1C0+n1C1.nC2+n1C2.nC3++n1Cn1.nCn

Consider (1+x)n=nC0+nC1x+nC2x2++nCnxn

1+1xn1=n1C0+n1C11x+n1C21x2++n1Cn11xn1

(1+x)n1+1xn1=[nC0+nC1x+nC2x2++nCnxn][n1C0+n1C11x+n1C21x2++n1Cn11xn1]

(1+x)2n1xn1=[nC0+nC1x++nCnxn] n1C0+n1C11x++n1Cn11xn1

Equating x coefficient on both sides

xncoefficient  in  (1+x)2n1xn1=n1C0nC1+n1C1nC2+

2n1Cn=n1C0.nC1+n1C1nC2+n1C0nC1+n1C1nC2+=2n1Cn=2n1C2n1n=2n1Cn1

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The value of  n−1C0. 1C0+ n−1C1. nC2+ n−1C2. nC3+⋯+ n−1Cn−1. nCn is equal to